What is the M.A.D of 1,1,1,1,5?

Sakura speaks 150 words per minute on average in Hungarian, and 190 words per minute on average in Polish. She once gave cooking

marusya05 [52]

Answer:3 minute

Step-by-step explanation:

Sakura speaks hungarian =150 words per minute

Sakura speaks polish =190 words per minute

and it is given she speaks 270 more words in polish than in hungarian

She speaks for a total of 5 minutes

let she speaks hungarian for t mins

therefore $150\times t+270=190\left ( 5-t\right )$

t=2 mins

therefore sakura speaks hungarian for 2 mins and polish for 3 mins

6 0

3 years ago

Someone help me with this plss

ladessa [460]

2.57 + 5x = 25.92 is the equation and when you solve for x you would get x = 4.67

3 0

2 years ago

Is this right ? Im honestly confused

fgiga [73]

Answer:

In step by step

Step-by-step explanation:

For #1, everything is perfect.

For #2, you wrote your functions out incorrectly, but your x input and y output are correct

#3 is a little messed up.

Your function is f(x)=-3x+1. Your given x's are -3, -1, 0, 1, 2, and 3. Plug all of these in.

1. f(-3)=-3(-3)+1

f(-3)=9+1
f(-3)=1

2. f(-1)=-3(-1)+1

f(-1)=3+1
f(-1)=4

3. f(0)=-3(0)+1

f(0)=0+1
f(0)=1

4. f(1)=-3(1)+1

f(1)=-3+1
f(1)=-3

5. f(2)=-3(2)+1

f(2)=-6+1
f(2)=-5

3 0

2 years ago

A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla

nekit [7.7K]

Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = $P(A)= \frac{13}{100}=0.13$

The probability that athlete is a basketball player = $P(B)= \frac{52}{100}=0.52$

The probability that athlete is both basket ball player and football player = $P(A\cap B) = \frac{9}{100}=0.09$

We have to find the probability that an athlete chosen is either a football player or a basketball player $P(A\cup B)$ .

Now we know that;

$P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%$

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

5 0

3 years ago

Algebra II :

Oduvanchick [21]

F(x) = 3x + 5/2
f(a+2) = 3(a+2) + 5/2 = 3a+6 + 5/2
= 3a + (12+5)/2 = 3a + 17/2

6 0

2 years ago